Difference between revisions of "1964 AHSME Problems/Problem 3"
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We can replace the letters with any numbers that satisfy this condition. It must be true for any numbers we use. Let <math>x</math> = <math>5</math> <math>y</math> = <math>2</math> <math>u</math> = <math>2</math> and <math>v</math> = <math>1</math>. Plug in you numbers and get <math>13</math> ÷ <math>2</math> <math>=</math> <math>6</math> remainder <math>1</math>. Since <math>v</math> = <math>1</math>, our answer is <math>E</math>. | We can replace the letters with any numbers that satisfy this condition. It must be true for any numbers we use. Let <math>x</math> = <math>5</math> <math>y</math> = <math>2</math> <math>u</math> = <math>2</math> and <math>v</math> = <math>1</math>. Plug in you numbers and get <math>13</math> ÷ <math>2</math> <math>=</math> <math>6</math> remainder <math>1</math>. Since <math>v</math> = <math>1</math>, our answer is <math>E</math>. | ||
| + | Solution by superagh | ||
Revision as of 17:04, 9 June 2019
We can replace the letters with any numbers that satisfy this condition. It must be true for any numbers we use. Let 
= 
= 
= and 
= 
. Plug in you numbers and get 
÷ 
remainder . Since = , our answer is 
.
Solution by superagh
